Press: RareBooksClub.com (May 11, 2012)
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1882 Excerpt: ...and the diameter of the circle described about the triangle.
Let ABC be a triangle, and let AD be the perpendicular from the angle A to the base BC: the rectangle BA, AC shall be equal to the rectangle contained by AD and the diameter of the circle described about the triangle.
Describe the circle ACB about the triangle; IV.
draw the diameter AE, and join EC.
Then, because the right angle BDA is equal to the angle ECA in a semicircle; III.
and the angle ABD is equal to the angle AEC, for they are in the same segment of the circle; PH.
therefore the triangle ABD ia equiangular to the triangle AEC.
Therefore BA is to AD as EA is to AC; VI: i.
therefore the rectangle BA, AC is equal to the rectangle EA, AD VI.
"Wherefore, if from the vertical angle &c Q.e.d.
THEOREM, The rectangle contained by the diagonals of a quadrilateral figure inscribed in a circle is equal to both the rectangles contained by its opposite sides..Let ABCD be any quadrilateral figure inscribed in a circle, and join AC,BD: the rectangle contained by A C, BD shall be equal to the two rectangles contained by AB, CD and by AD, BC.
Make the angle ABE equal to the angle DBC; I.
add to each of these equals the angle EBD, then the angle ABD is equal to the angle EBC.
And the angle BDA is equal to the angle B CE, for they are in the same segment of the circle; 111.21.
therefore the triangle A BD is equiangular to the triangle EBG.
Therefore A D is to DB as EC is to CB; therefore the rectangle AD, CB is equal to the rectangle DB, EC.
Again, because the angle ABE is equal to the angle DBC, Construction.
and the angle BAE is equal to the angle BDC, for they are in the same segment of the circle; III.
therefore the triangle ABE is equiangu...
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